Related to Constant acceleration

[HelpMeHelpYou]

Homework Statement

Suppose a child driving a go-cart is traveling 4m/s when she crosses a line 4m from her starting point. She continues with a steady acceleration of 0.4m/s^2 , until she crosses a mark 40m from the starting point. How long does it take for her to go from the 4m mark to the 40m mark?


what i did is this :

a = 0.4m/s^2
Vi = 4m/s
d = 40m - 4m = 36m

i used this formula :


Vf = Vi + a*t
0 = 4m/s + (0.4m/s^2)t
-0.4m/s^2 = 4m/s
t = -10s

and this :

d = Vi+Vt/2 * t
36m = (4m/s + 0)/2 * t
36m = 2m/s t

divide both side by 2m/s

t = 18s


but the true answer is 6.7s

anyone correct this ?

 

[sambristol]

the direct way to solve it, without having to calculate the final velocity, is to use
d = V₁ t + a t² /2
put in the values d= 36, a=0.4 and solve the quadratic equation in t using the normal formula. Its obviously the positive value of t you take.

Hope this helps

Regards

Sam

 

[HelpMeHelpYou]

how about the velocity ? Vi = 0 or 4m/s ?

 

[sambristol]

4 m/s

 

[andrien]

36=4t+0.2t^2

 

[HelpMeHelpYou]

just a confirmation ..

correct this if I'm wrong ..

d = 36
a = 0.4

i used :

d = V1 t + a t2 /2
subst:

36 = 0 t + 0.4 t2 /2
divide both sides by 0.4

90 = t2 /2
divide both sides by 1/2

45 = t2
square root both sides to remain t :

t = 3√5 or 6.7

is it correct usage of algebra ?

 

[sambristol]

try 36 = 4t+0.2t²

0.2 t² +4t -36 =0

t = (-4 + √(16+4*0.2*36))/(2*0.2)

t≈6.7 seconds

 

[HelpMeHelpYou]

whow ! again .. i should study back again .. so thanks Sam ..

gratitude !